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Math Help - Partial Fractions

  1. #1
    Junior Member
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    Angry Partial Fractions

    The question simply ask to write the partial fraction decomposition of the rational expression.

    (x^2) / (x^4 - 2x^2 - 8)

    I wrote it out as:

    (A / (x-2)) + (B / (x+2)) + ((Cx+D) / (x^2+2))

    I got as far as simplifying it as:

    Ax^3+2Ax^2+2Ax+4A+Bx^3-2Bx^2+2Bx-4B+Cx^3-4Cx+Dx^2-4D

    So I group them up as:

    (A+B+C)x^3=0
    (2A-2B+D)x^2=1
    (2A+2B-4C)x=0
    (A-B-D)4=0

    Now I'm lost in using the elimination method because I just don't see how to peace together this puzzle, and the use the answer to complete the partial fraction decomposition. OMG I need some assistance please, this problem is driving me mad...
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  2. #2
    MHF Contributor
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    Quote Originally Posted by ugkwan View Post
    The question simply ask to write the partial fraction decomposition of the rational expression.

    (x^2) / (x^4 - 2x^2 - 8)

    I wrote it out as:

    (A / (x-2)) + (B / (x+2)) + ((Cx+D) / (x^2+2))

    I got as far as simplifying it as:

    Ax^3+2Ax^2+2Ax+4A+Bx^3-2Bx^2+2Bx-4B+Cx^3-4Cx+Dx^2-4D

    So I group them up as:

    (A+B+C)x^3]=0
    (2A-2B+D)x^2=1
    (2A+2B-4C)]x=0
    (A-B-D)4=0

    Now I'm lost in using the elimination method because I just don't see how to peace together this puzzle, and the use the answer to complete the partial fraction decomposition. OMG I need some assistance please, this problem is driving me mad...
    hi ugkwan,

    continuing on, this means

    A+B+C=0 (1)
    2A-2B+D=1 (2)
    2A+2B-4C=0 (3)
    A-B-D=0 (4)

    This is a system of 4 simultaneous equations,
    for which there is enough information to solve for the 4 values A, B, C, D

    Combining (1) and (3)....2A+2B+2C=0 and 2A+2B-4C=0

    subtracting these gives 6C=0, so C=0

    Hence 2A+2B=0 gives A=-B

    Hence, from (4) we have 2A-D=0, so 2A=D

    therefore, using (2).....2A-2B+D=1 gives 4A+D=1 so 2D+D=1 so D=1/3

    Hence A=1/6 and B=-1/6
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